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Let $(A,\,^\ast,\lVert\cdot\rVert)$ be a Banach $\ast$-algebra with the property that $\lVert a \rVert^2=\rho(a^\ast a)$ holds for all $a\in A$, where $\rho(x)$ denotes the spectral radius of an element $x\in A$.

Is $(A,\,^\ast,\lVert\cdot\rVert)$ then a $C^\ast$-algebra?

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[Updated to include Nik Weaver's correction / improvement from the comments.]

I think this follows from the spectral radius formula:

$$\left\| a \right\|^2 = \rho(a^*a) = \lim_{k \to \infty} \left\| (a^*a)^k \right\|^{1/k} \leq \left\| a^*a \right\| \leq \left\| a^* \right\| \left\| a \right\|$$

This gives $\left\| a \right\|^2 \leq \left\| a^* \right\| \left\| a \right\|$, implying $\left\| a \right\| \leq \left\| a^* \right\|$. Applying the same argument to $a^*$ we get $\left\| a \right\| = \left\| a^* \right\|$, so the inequality above becomes:

$$\left\| a \right\|^2 \leq \left\| a^* a \right\| \leq \left\| a \right\|^2$$

which yields the C* identity, as desired.

(The inequalities above follow from the Banach algebra axiom $\left\| ab \right\| \leq \left\| a \right\| \left\| b \right\|$.)

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    $\begingroup$ I think you need to say a little more ... first use $\|a^*a\| \leq \|a^*\|\|a\|$ in your displayed line to get $\|a\| \leq \|a^*\|$, then $\|a\| = \|a^*\|$ by symmetry, so $\|a^*a\| \leq \|a\|^2$. $\endgroup$
    – Nik Weaver
    Commented Mar 4, 2020 at 20:32
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    $\begingroup$ (Unless you count $\|a\| = \|a^*\|$ as an axiom for "Banach *-algebra" ...) $\endgroup$
    – Nik Weaver
    Commented Mar 4, 2020 at 20:33
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    $\begingroup$ Alternatively, if you define the spectral radius as the largest modulus of an element of the spectrum, then $\rho(x)\leq \Vert x\Vert$ by the Neumann lemma and so you can bypass the BG-spectral radius formula $\endgroup$
    – Yemon Choi
    Commented Mar 4, 2020 at 21:19
  • $\begingroup$ @NikWeaver You're right - I shouldn't have used $\left\| a^* \right\| = \left\| a \right\|$ without comment. I wouldn't consider it to be a Banach *-algebra axiom, but your correction fixes the argument anyway. $\endgroup$
    – Paul Siegel
    Commented Mar 5, 2020 at 3:31
  • $\begingroup$ @PaulSiegel: I googled it, and found that according to Wikipedia "Some authors include this isometric property in the definition of a Banach *-algebra." $\endgroup$
    – Nik Weaver
    Commented Mar 5, 2020 at 3:35

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